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If a, b, and c are integers and \(a \lt b \lt c\), are a, b, and c consecutive integers? (1) The median of {a!, b!, c!} is an odd number. (2) c! is a prime number.
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Re D0120 [#permalink]
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15 Sep 2014, 23:12
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Official Solution: Note that: A. The factorial of a negative number is undefined. B. 0!=1. C. Only two factorials are odd: 0!=1 and 1!=1. D. Factorial of a number which is prime is 2!=2. (1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient. (2) c! is a prime number. This implies that c=2. Not sufficient. (1)+(2) From above we have that a=0, b=1 and c=2, thus the answer to the question is YES. Sufficient. Answer: C
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Re: D0120 [#permalink]
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23 Oct 2014, 03:22
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Bunuel wrote: If a, b, and c are integers and \(a \lt b \lt c\), are a, b, and c consecutive integers?
(1) The median of {a!, b!, c!} is an odd number.
(2) c! is a prime number. I got this question wrong in the test, because I didn't pay special attention to every detail given in the question. Here's the right solution : 1. The median of a! b! c! is an odd number, it means that b is either 0 or 1. Since we can have a!, it means that a is greater than or equal to 0. We also know that a < b < c, so a = 0, b = 1, but we have no information about c. SO, this is insufficient. 2. c! is a prime number, this means that c = 2. Here we have no information about a and b, so Insufficient. Combining 1 & 2, we get that a = 0, b = 1, c=2. So, C
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Re: D0120 [#permalink]
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17 Dec 2014, 04:37
Statement 1: According to the statement,median of a!,b!,c! is a odd number. Only the factorial of integer 1 is a odd integer.Hence b should be 1 but c can be any value greater than 1. Hence not sufficient.
Statement 2: Only information about c.Not sufficient.
Combining both statements: Given c! is a prime number and b=1 then,the only 2! gives a prime number 2(only even prime). The factorial of other integers are not prime. Hence a=0,b=1,c=2. Answer:C



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Re: D0120 [#permalink]
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19 Dec 2014, 01:13
Really a very nice question !! +1 for Bunuel(s).
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Re: D0120 [#permalink]
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11 Jan 2015, 09:49
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I think this question is poor and helpful. I believe the given answer is incorrect and the correct one is B > statement 2 by itself is sufficient. Indeed, if c! is prime then it must be 2 and since a < b < c and a,b,c are INTEGERS AND factorial is not defined for negative numbers then a and b have no other choice but to be 0 and 1 respectively! Thus, statement 2 is enough by itself. I would like to hear your comments. Thanks.
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Re: D0120 [#permalink]
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11 Jan 2015, 10:18
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Re: D0120 [#permalink]
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28 Jan 2015, 10:29
I think this question is good and helpful. I think answer should be "Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient." because if c=2 then a=0 and b=1 follows.



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Re: D0120 [#permalink]
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29 Jan 2015, 04:21



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Re: D0120 [#permalink]
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18 Aug 2015, 23:28
I think this is a highquality question and I agree with explanation. Really good question



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Re D0120 [#permalink]
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20 Feb 2016, 22:30
I don't agree with the explanation. a, b, c are consecutive integers.
So, if we agree that b can only be 1 from statement 1, then, c has to be 2.



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Re: D0120 [#permalink]
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22 Feb 2016, 00:15
bhaskarj4 wrote: I don't agree with the explanation. a, b, c are consecutive integers.
So, if we agree that b can only be 1 from statement 1, then, c has to be 2. We do NOT know that a, b, and c are consecutive integers. The question is: are a, b, and c consecutive integers?
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Re D0120 [#permalink]
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26 Jun 2016, 12:18
I think this is a highquality question and I don't agree with the explanation. Statement 2: If we know that only value for c is 2 and from Question we know a<b<c Additionally ve factorial is undefined, then only option we have for a, b, and c is 0, 1, 2.
Hence this should be sufficient. Please help explain if this analysis is correct.



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Re: D0120 [#permalink]
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26 Jun 2016, 12:21



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Re D0120 [#permalink]
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06 Jul 2016, 07:32
I think this is a highquality question and I agree with explanation. Terrific Trick! Bunuel the Man



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Re D0120 [#permalink]
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12 Jul 2016, 11:17
I think this is a highquality question and I agree with explanation.



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Re: D0120 [#permalink]
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21 Jul 2016, 21:22
Bunuel and others: I have similar doubt like others. I appreciate your inputs. Lets ignore Stmt A for now. Question says, If a, b, and c are integers and a<b<c, are a, b, and c consecutive integers? Stmt(1): ignore. Stmt(2): c! is a prime number. only possible number is 2. !2 =2 a prime number. Now, a<b<c and they are integers. The only possible combination i see here is 0,1,2. I ignored combination with negative numbers such as 1,1,2 because !negative is undefined. Where am i going wrong here?
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Re: D0120 [#permalink]
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21 Jul 2016, 21:27
JarvisR wrote: Bunuel and others: I have similar doubt like others. I appreciate your inputs. Lets ignore Stmt A for now. Question says, If a, b, and c are integers and a<b<c, are a, b, and c consecutive integers? Stmt(1): ignore. Stmt(2): c! is a prime number. only possible number is 2. !2 =2 a prime number. Now, a<b<c and they are integers. The only possible combination i see here is 0,1,2. I ignored combination with negative numbers such as 1,1,2 because !negative is undefined. Where am i going wrong here? You do not have factorial of a or b in stem or in (2) so you cannot ignore negative numbers.
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Re: D0120 [#permalink]
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21 Jul 2016, 21:33
Ahhh... now this is embarrassing. Sorry, in my stupidity i missed that part.
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13 Sep 2016, 19:05
I think this is a highquality question and I agree with explanation.







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